1）通过电脑和打印机的倍数关系确定二者大小关系，电脑C最多是打印机P的3倍，也就是C ≤ 3P
The number of computers can be at most 3 times the number of printers
Let C = the number of computers purchased
Let P = the number of printers purchased
So, we can write: C ≤ 3P
Subtract 3P from both sides of the inequality to get: C - 3P ≤ 0

2）买电脑和打印机一起的钱z最多不超过是$9100
Computers cost $1,500 each, and printers cost $300 each.
So, the total cost of purchasing C computers and P printers = 1500C + 300P

What is the greatest number of computers can the company buy if it has a total of $9,100 to spend on computers and printers?
This tells us that the total cost cannot exceed $9100
So, we can write, 1500C + 300P ≤ 9100

3）联立不等式，求得C的范围
We now have the following system of inequalities:
C - 3P ≤ 0
1500C + 300P ≤ 9100

Take the top inequality and multiply both sides by 100 to get:
100C - 300P ≤ 0
1500C + 300P ≤ 9100

Since the inequality symbols are facing the same direction, we can add the two inequalities to get: 1600C ≤ 9100
Divide both sides of the inequality by 1600 to get: C ≤ 9100/1600
Simplify to get: C ≤ 5.something

C最大是5